codeforces 868C

Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of n problems, and they want to select any non-empty subset of it as a problemset.

k experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.

Determine if Snark and Philip can make an interesting problemset!

Input

The first line contains two integers n, k (1 ≤ n ≤ 10⁵, 1 ≤ k ≤ 4) — the number of problems and the number of experienced teams.

Each of the next n lines contains k integers, each equal to 0 or 1. The j-th number in the i-th line is 1 if j-th team knows i-th problem and 0 otherwise.

Output

Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.

You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").

Examples

Input

Output

NO

Input

Output

YES

Note

In the first example you can't make any interesting problemset, because the first team knows all problems.

In the second example you can choose the first and the third problems.

这题比赛时1a了，当时还挺开心，觉得自己暴力写了个O(N)的代码。结果结束后看人家的ac代码，顿时感觉自己是个zz。

我在想思路的时候，已经用四位数（题目指出最多四个队伍，我们可以假设一直都有四个队伍，也是不影响的）来表示了，但是还是没想到用二进制。

最后只能一点一点的用if来判断。

附我的ac代码：

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <iostream>
  4 #include <string>
  5 #include <algorithm>
  6 #include <cmath>
  7 using namespace std;
  8 const int maxn = 100005;
  9 int nu[maxn][6];
 10 int cnt[22];
 11 int main() {
 12     ios::sync_with_stdio(false);
 13     int n,k;
 14     cin>>n>>k;
 15     for(int i = 0; i < n; i++) {
 16         for(int j = 0; j < k; j++) {
 17             cin>>nu[i][j];
 18         }
 19     }
 20     for(int i = 0; i < n; i++) {
 21         if(nu[i][0] == 0 ) {
 22             if(nu[i][1] == 0) {
 23                 if(nu[i][2] == 0) {
 24                     if(nu[i][3] == 0) {
 25                         cnt[16]++;
 26                     }
 27                     else {
 28                         cnt[1]++;
 29                     }
 30                 }
 31                 else {
 32                     if(nu[i][3] == 0) {
 33                         cnt[2]++;
 34                     }
 35                     else{
 36                         cnt[10]++;
 37                     }
 38                 }
 39             }
 40             else {
 41                 if(nu[i][2] == 0) {
 42                     if(nu[i][3] == 0) {
 43                         cnt[3]++;
 44                     }
 45                     else {
 46                         cnt[9]++;
 47                     }
 48                 }
 49                 else {
 50                     if(nu[i][3] == 0) {
 51                         cnt[8]++;
 52                     }
 53                     else{
 54                         cnt[14]++;
 55                     }
 56                 }
 57             }
 58         }
 59         else {
 60             if(nu[i][1] == 0) {
 61                 if(nu[i][2] == 0) {
 62                     if(nu[i][3] == 0) {
 63                         cnt[4]++;
 64                     }
 65                     else {
 66                         cnt[7]++;
 67                     }
 68                 }
 69                 else {
 70                     if(nu[i][3] == 0) {
 71                         cnt[6]++;
 72                     }
 73                     else{
 74                         cnt[13]++;
 75                     }
 76                 }
 77             }
 78             else {
 79                 if(nu[i][2] == 0) {
 80                     if(nu[i][3] == 0) {
 81                         cnt[5]++;
 82                     }
 83                     else {
 84                         cnt[12]++;
 85                     }
 86                 }
 87                 else {
 88                     if(nu[i][3] == 0) {
 89                         cnt[11]++;
 90                     }
 91                     else{
 92                         cnt[15]++;
 93                     }
 94                 }
 95             }
 96         }
 97     }
 98     if(cnt[16]) cout<<"YES"<<endl;
 99     else if(cnt[1]) {
100         if(cnt[2]||cnt[3]||cnt[4]||cnt[5]||cnt[6]||cnt[8]||cnt[11])
101         cout<<"YES"<<endl;
102         else cout<<"NO"<<endl;
103     }
104     else if(cnt[2]) {
105         if(cnt[3]||cnt[4]||cnt[5]||cnt[7]||cnt[9]||cnt[12])
106         cout<<"YES"<<endl;
107         else cout<<"NO"<<endl;
108     }
109     else if(cnt[3]) {
110         if(cnt[4]||cnt[6]||cnt[7]||cnt[10]||cnt[13])
111         cout<<"YES"<<endl;
112         else cout<<"NO"<<endl;
113     }
114     else if(cnt[4]) {
115         if(cnt[8]||cnt[9]||cnt[10]||cnt[14])
116         cout<<"YES"<<endl;
117         else cout<<"NO"<<endl;
118     }
119     else if(cnt[5]) {
120         if(cnt[10])
121         cout<<"YES"<<endl;
122         else cout<<"NO"<<endl;
123     }
124     else if(cnt[6]) {
125         if(cnt[9])
126         cout<<"YES"<<endl;
127         else cout<<"NO"<<endl;
128     }
129     else if(cnt[7]) {
130         if(cnt[8])
131         cout<<"YES"<<endl;
132         else cout<<"NO"<<endl;
133     }
134     else cout<<"NO"<<endl;
135     
136     return 0;
137 }

View Code

附别人的ac代码：

 1 #include<cstdio>
 2 int n,k,a[17],b[5];
 3 int main(){
 4     scanf("%d%d",&n,&k);
 5     while(n--){
 6         for(int i=1;i<=k;i++)scanf("%d",&b[i]);
 7         a[b[1]+b[2]*2+b[3]*4+b[4]*8]=1;  //化为二进制 
 8     }
 9     for(int i=0;i<=(1<<k);i++)
10         for(int j=i;j<=(1<<k);j++)
11             if(!(i&j)&&a[i]&&a[j])  //如果存在i和j四位都不同时为1 例：1010 0100 
12                 {puts("YES");return 0;} 
13     puts("NO");
14     return 0;
15 }

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